contains

Declaration: let listArray = ["kashif"] let word = "kashif" then this contains(listArray, word) Returns true but if declaration is: let word = "Kashif" then it returns false because comparison is case sensitive. How to make this comparison case insensitive? you can use word.lowercaseString to convert the string to all lowercase characters let list = ["kashif"] let word = "Kashif" if contains(list, {$0.caseInsensitiveCompare(word) == NSComparisonResult.OrderedSame}) { println(true) // true } Xcode 7.3.1 • Swift 2.2.1 if list.contains({$0.caseInsensitiveCompare(word) == .OrderedSame}) { print

Is it possible to have the contain function find if the string contains 2 words or more? This is what I'm trying to do: string d = "You hit someone for 50 damage"; string a = "damage"; string b = "someone"; string c = "you"; if(d.Contains(b + a)) { Console.WriteLine(" " + d); Console.ReadLine(); } When I run this, the console window just shuts down really fast without showing anything. And another question: if I for one want to add how much damage is done, what would be the easiest way to get that number and get it into a TryParse ? You would be better off just calling Contains twice or making

I am trying to check if a specific item (value of a property) exists in a array of objects, but could not find out any solution. Please let me know, what i am missing here. class Name { var id : Int var name : String init(id:Int, name:String){ self.id = id self.name = name } } var objarray = [Name]() objarray.append(Name(id: 1, name: "Nuibb")) objarray.append(Name(id: 2, name: "Smith")) objarray.append(Name(id: 3, name: "Pollock")) objarray.append(Name(id: 4, name: "James")) objarray.append(Name(id: 5, name: "Farni")) objarray.append(Name(id: 6, name: "Kuni")) if contains(objarray["id"], 1) {

Basically I have about 1,000,000 strings, for each request I have to check if a String belongs to the list or not. I'm worried about the performance, so what's the best method? ArrayList ? Hash? krock Your best bet is to use a HashSet and check if a string exists in the set via the contains() method. HashSets are built for fast access via the use of Object methods hashCode() and equals() . The Javadoc for HashSet states: This class offers constant time performance for the basic operations (add, remove, contains and size), HashSet stores objects in hash buckets which is to say that the value

My problem is that i want to check the browserstring with pure regex. Mozilla/5.0 (Linux; U; Android 3.0; en-us; Xoom Build/HRI39) AppleWebKit/534.13 (KHTML, like Gecko) Version/4.0 Safari/534.13 -> should match Mozilla/5.0 (Linux; U; Android 2.2.1; en-us; Nexus One Build/FRG83) AppleWebKit/533.1 (KHTML, like Gecko) Version/4.0 Mobile Safari/533.1 should not match my tried solution is: /?((?<=Android)(?:[^])*?(?=Mobile))/i but it matches exactly wrong. You use look ahead assertions to check if a string contains a word or not. If you want to assure that the string contains "Android" at some

I've done some searching and can't figure out how to filter a dataframe by df["col"].str.contains(word) , however I'm wondering if there is a way to do the reverse: filter a dataframe by that set's compliment. eg: to the effect of !(df["col"].str.contains(word)) . Can this be done through a DataFrame method? Andy Hayden You can use the invert (~) operator (which acts like a not for boolean data): new_df = df[~df["col"].str.contains(word)] , where new_df is the copy returned by RHS. contains also accepts a regular expression... If the above throws a ValueError, the reason is likely because you